思路：一个篱笆的作用就是堵死一条路，我们将S与狼建边，权值为inf， 将羊与T建边，权值为inf， 每个点与其可以到的路建边，权值为1，

然后跑最小割，因为如果S到T有通路说明狼肯定可以到羊，每一个割代表一个篱笆。

#include
     
      #define LL long long#define fi first#define se second#define mk make_pair#define pii pair
      
       using namespace std;const int N = 2e4 + 7;const int M = 2e5 + 7;const int inf = 0x3f3f3f3f;const LL INF = 0x3f3f3f3f3f3f3f3f;const int mod = 1e9 + 7;int n, m, S, T, tot, level[N], head[N], a[101][101], id[101][101];int dx[] = {
   0, 0, 1, -1}, dy[] = {
   1, -1, 0, 0};struct Edge {    int to, w, nx;}edge[M << 1];void add(int u, int v, int w) {    edge[tot].to = v;    edge[tot].w = w;    edge[tot].nx = head[u];    head[u] = tot++;    edge[tot].to = u;    edge[tot].w = 0;    edge[tot].nx = head[v];    head[v] = tot++;}bool bfs() {    memset(level, 0, sizeof(level));    queue
       
         que;    que.push(S), level[S] = 1;    while(!que.empty()) {        int u = que.front(); que.pop();        if(u == T) return true;        for(int i = head[u]; ~i; i = edge[i].nx) {            int v = edge[i].to, w = edge[i].w;            if(level[v] || w <= 0) continue;            level[v] = level[u] + 1;            que.push(v);        }    }    return false;}int dfs(int u, int p) {    if(u == T) return p;    int ret = 0;    for(int i = head[u]; ~i; i = edge[i].nx) {        int v = edge[i].to, w = edge[i].w;        if(level[v] != level[u] + 1 || w <= 0) continue;        int f = dfs(v, min(p - ret, w));        ret += f;        edge[i].w -= f;        edge[i ^ 1].w += f;        if(ret == p) break;    }    if(!ret) level[u] = 1;    return ret;}int Dinic() {    int ans = 0;    while(bfs()) ans += dfs(S, inf);    return ans;}void init() {    tot = 0;    memset(head, -1, sizeof(head));}bool check(int x, int y) {    if(x < 1 || x > n) return false;    if(y < 1 || y > m) return false;    return true;}int main() {    init();    scanf("%d%d", &n, &m);    S = 0, T = n * m + 1;    for(int i = 1; i <= n; i++) {        for(int j = 1; j <= m; j++) {            scanf("%d", &a[i][j]);            id[i][j] = (i - 1) * m + j;            if(a[i][j] == 1) add(S, id[i][j], inf);            else if(a[i][j] == 2) add(id[i][j], T, inf);        }    }    for(int i = 1; i <= n; i++) {        for(int j = 1; j <= m; j++) {            for(int k = 0; k < 4; k++) {                int x = i + dx[k], y = j + dy[k];                if(check(x, y) && (a[i][j] != a[x][y] || !a[i][j] && !a[x][y])) {                    add(id[i][j], id[x][y], 1);                }            }        }    }    printf("%d\n", Dinic());    return 0;}/**/